# Example: Comparison of Bi-elliptic and Two-Impulse Hohmann Transfers¶

As we discussed, bi-elliptic transfers can save $$\Delta v$$ requirement when the initial and target orbits have very different radii. When the target orbit radius is more than about 15.5 times larger than the initial radius (or vice versa), the bi-elliptic transfer is more energy efficient than the standard, two-impulse, Hohmann transfer.

In this example, we will explore this with some numbers, and also discuss the tradeoffs of a bielliptic transfer.

Let’s start the initial orbit at the radius of the moon, 385,000 km, and set the target as an LEO orbit of 500 km altitude. Assuming both orbits are circular, we can find the initial and target orbital velocities using Eq. (121):

import numpy as np

mu = 398_600.4418  # km**3/s**2
R_E = 6378  # km
r_1 = 385_000  # km

v_1 = np.sqrt(mu / r_1)  # km/s

r_3 = 500 + R_E  # km
v_3 = np.sqrt(mu / r_3)  # km/s


This gives velocities of $$v_1 =$$ 1.018 km/s and $$v_3 =$$ 7.613 km/s.

Now we will calculate the standard two-impulse Hohmann transfer. The total $$\Delta v$$ for the transfer is the sum of the $$\Delta v$$ from the initial orbit onto the transfer orbit at perigee of the transfer, and the $$\Delta v$$ from the transfer orbit onto the target orbit at the apogee of the transfer.

First, we calculate the eccentricity, specific orbital angular momentum, and semimajor axis of the transfer orbit:

$e_t = \frac{r_1 - r_3}{r_3 + r_1}$
$h_t = \sqrt{\mu r_1 (1 - e)} = \sqrt{\mu r_3 (1 + e)}$
$a_t = \frac{r_3 + r_1}{2}$
e_t = (r_1 - r_3) / (r_1 + r_3)
h_t = np.sqrt(r_1 * (1 - e_t) * mu)
a_t = (r_3 + r_1) / 2


This gives $$e_t =$$ 0.9649, $$h_t =$$ 73395.578 km2/s, and $$a_t =$$ 1.959e+05 km. Now, the velocity at the perigee and apogee of the transfer orbit are determined from the specific orbital angular momentum.

v_p_t = h_t / r_3
v_a_t = h_t / r_1


The velocity at the apogee of the transfer orbit is $$v_{a,t} =$$ 0.191 km/s, but the orbital velocity at the moon’s orbit is over 1 km/s, a factor of nearly five. Now, let’s calculate $$\Delta v$$:

delta_v_t = abs(v_p_t - v_3) + abs(v_1 - v_a_t)


The $$\Delta v$$ for the two-impulse Hohmann transfer is $$\Delta v =$$ 3.885 km/s.

Next, let’s do the bielliptic transfer. We pick $$r_2 =$$ 770,000 km, that is, apogee of the transfer orbit is 770,000 km from the center of the Earth. Now, we can calculate the eccentricity, specific orbital angular momentum, and velocities for the two transfer ellipses.

r_2 = 770_000  # km
e_t1 = (r_2 - r_1) / (r_1 + r_2)
h_t1 = np.sqrt(r_2 * (1 - e_t1) * mu)
v_p_t1 = h_t1 / r_1
v_a_t1 = h_t1 / r_2

e_t2 = (r_2 - r_3) / (r_3 + r_2)
h_t2 = np.sqrt(r_2 * (1 - e_t2) * mu)
v_p_t2 = h_t2 / r_3
v_a_t2 = h_t2 / r_2


These velocities are $$v_{p,t1} =$$ 1.175 km/s, $$v_{a,t1} =$$ 0.587 km/s, $$v_{p,t2} =$$ 10.718 km/s, and $$v_{a,t2} =$$ 0.096 km/s.

The difference in apogee velocities between the two transfer orbits is about 0.5 km/s, and now the velocity difference at lunar orbit is only about 0.18 km/s, much lower than before. The total $$\Delta v$$ for these three impulses is:

delta_v_b = abs(v_p_t1 - v_1) + abs(v_a_t2 - v_a_t1) + abs(v_p_t2 - v_3)
delta_v_diff = abs(delta_v_t - delta_v_b) * 1000  # m/s


For the bi-elliptic transfer, $$\Delta v =$$ 3.755 km/s, about 131 m/s, or 3.36% less than the two-impulse Hohmann transfer.

Assuming a 1000 kg spacecraft with an $$I_{sp}$$ of 300 s, this results in a savings of propellant of:

delta_m_t = 1000 * (1 - np.exp(-delta_v_t / (300 * 9.81E-3)))
delta_m_b = 1000 * (1 - np.exp(- delta_v_b / (300 * 9.81E-3)))
delta_m_diff = abs(delta_m_t - delta_m_b)


The difference is 12.1 kg per 1000 kg of spacecraft mass.

Although those numbers seem small, let’s put them in context. For the Falcon 9, the Full Thrust variant has a mass of 549,000 kg. The savings from the bi-elliptic transfer means that about 7,000 kg of fuel can be diverted to another use. The total payload capacity to Low Earth Orbit is about 23,000 kg, so this is a significant savings. Although this is a simplistic model of the rocket, we can at least see the order of magnitude of savings that are possible.

On the other hand, the downside of the bielliptic transfer is the transit time. The period of an ellipse is given by Eq. (138).

t_h = np.pi / np.sqrt(mu) * a_t**(3/2)
a_be_1 = (r_2 + r_1) / 2
a_be_2 = (r_2 + r_3) / 2
t_be = np.pi / np.sqrt(mu) * (a_be_1**(3/2) + a_be_2**(3/2))


The transfer time for the two-impulse Hohmann transfer is $$t_h =$$ 4.995 days, and for the bi-elliptic transfer it is $$t_{be} =$$ 39.218 days. The transfer time is almost 8x longer for the bielliptic transfer!